Question

c++

Please give a line by line explanation fo the design of this code (what is happening per line to make this program work)

int ways(int amt,int* l,int n,int* res){
if (amt < 0 || (amt > 0 && n <= 0)) return 0;
if (amt == 0){
string s = "";
int j = 0;
if (res[0] > 0){
j = 1;

cout << res[0] << " quarter/s, ";
}
if (res[1] > 0) {
j = 1;
cout << res[1] << " dime/s, ";
}
if (res[2] > 0){
j = 1;
cout << res[2] << " nickel/s, ";
}
if (res[3] > 0){
if (j == 1) cout << "and " <<res[3] << " pennies";
else cout << res[3] << " pennies";
}
cout << endl;
return 1;
}
int* temp = new int[4];
for (int i = 0; i < 4; i++)
temp[i] = res[i];
int x = ways(amt,l,n-1,temp);
temp[n-1] += 1;
int y = ways(amt-l[n-1],l,n,temp);
return x+y;
}

int main(){
int amt;
cout << "Enter the amount of pennies: ";
cin >> amt;
cout << "These are the combinations: " "\n";
int* l = new int[4];
l[0] = 25; l[1] = 10; l[2] = 5; l[3] = 1;

int* res = new int[4];
res[0] = 0; res[1] = 0; res[2] = 0; res[3] = 0;
cout << endl << "The total amount of different combinations: " << ways(amt,l,4,res) << endl;
return 0;
}

Answer 1

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Info:

1 nickel = 5 pennies

1 dime = 10 pennies

1 quarter = 25 pennies

Description of program:

The program has a recursive function called ways which are calculating the combination in recursive ways. In basic terms, when you enter the amount of pennies as input then the program print the combinations in which the given pennies can be divided into nickel, dime and quarters and also the count of those combinations. (Combinations are the ways to group numbers).

CODE:

#include <iostream>
#include <string>
using namespace std;

int ways(int amt,int* l,int n,int* res){
  
//if the amount is 0 or the size of array l is 0, function will return 0
if (amt < 0 || (amt > 0 && n <= 0))
return 0;
  
// When the amount becomes zero or we can say that this program will enter in this condition for every recursive call
// where the current amount becomes zero.
// Also, when in a recursive call, the current amount becomes 0, the function uses the res value which is the current value
// of current temp pointer array.
// The if condition checks the index of res if any index value of res is > 0 then that index value is printed with a unit string
// After printing, the program returns 1 which is then added to the count of the combination of pennies.
if (amt == 0){
  
   // This string variable is not used anywhere in the program
string s = "";
  
   // This variable is used to use in the last if condition, to check whether above if conditions are executed or not
   // that means, quarter's, dime's or nickel are possible for the given amount.
int j = 0;
  
   //First if condition to check if quarters are possible for the given amount
if (res[0] > 0){
j = 1;
cout << res[0] << " quarter/s, ";
}

   //Second if condition to check if dime's are possible for the given amount
if (res[1] > 0) {
j = 1;
cout << res[1] << " dime/s, ";
}

   //Third if condition to check if nickel's are possible for the given amount
if (res[2] > 0){
j = 1;
cout << res[2] << " nickel/s, ";
}

   //Last if condition to check the remaining pennies
if (res[3] > 0){
  
   //When above if conditions are true, then it will print " and ___ pennies"
if (j == 1)
cout << "and " <<res[3] << " pennies";
   //When above if conditions are true, then it will print " ___ pennies"
else
cout << res[3] << " pennies";
}
  
   //For ending the current line
   cout << endl;

   //return statement
return 1;
}
  
// Creating a pointer of arrays temp
int* temp = new int[4];
  
// Copying the res array to temp array
for (int i = 0; i < 4; i++)
temp[i] = res[i];
  
// This variable runs, for the count of n
int x = ways(amt,l,n-1,temp);
  
// Every time the function is called when amount is non-zero, the value of index of temp is increased by 1
temp[n-1] += 1;
  
// This variable checks whether the current amount can be converted to any unit of a dollar
// amt - l[n-1] subtracts the current amount with the unit amount like a quarter, dime, nickel etc, and checks
// if it current amount is equal to or greater than the unit
int y = ways(amt-l[n-1],l,n,temp);
  
// returning the sum of x+y
return x+y;
}

// Main function
int main(){
  
int amt;

cout << "Enter the amount of pennies: ";
cin >> amt;
  
cout << "These are the combinations: " "\n";
// pointer array which stores the amounts of units of dollar
int* l = new int[4];
l[0] = 25; l[1] = 10; l[2] = 5; l[3] = 1;
  
// Another pointer array which is used in way() to store the count of units in every recursive call.
int* res = new int[4];
res[0] = 0; res[1] = 0; res[2] = 0; res[3] = 0;

cout << endl << "The total amount of different combinations: " << ways(amt,l,4,res) << endl;
  
return 0;
}

// Code ends here!!!!!!!!!!

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